Problem: Divide the following complex numbers. $ \dfrac{4+2i}{-1+i}$
Solution: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-1-i}$ $ \dfrac{4+2i}{-1+i} = \dfrac{4+2i}{-1+i} \cdot \dfrac{{-1-i}}{{-1-i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(4+2i) \cdot (-1-i)} {(-1+i) \cdot (-1-i)} = \dfrac{(4+2i) \cdot (-1-i)} {(-1)^2 - (1i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(4+2i) \cdot (-1-i)} {(-1)^2 - (1i)^2} = $ $ \dfrac{(4+2i) \cdot (-1-i)} {1 + 1} = $ $ \dfrac{(4+2i) \cdot (-1-i)} {2} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({4+2i}) \cdot ({-1-i})} {2} = $ $ \dfrac{{4} \cdot {(-1)} + {2} \cdot {(-1) i} + {4} \cdot {-1 i} + {2} \cdot {-1 i^2}} {2} $ Evaluate each product of two numbers. $ \dfrac{-4 - 2i - 4i - 2 i^2} {2} $ Finally, simplify the fraction. $ \dfrac{-4 - 2i - 4i + 2} {2} = \dfrac{-2 - 6i} {2} = -1-3i $